What are the solutions to the equation x^2+20x+68=0 ?

The solutions to the equation x^2+20x+68=0 are x=2(2sqrt(2)-5),x=-2(5+2sqrt(2))

Find the zeros of x^2+20x+68=0

The zeros of x^2+20x+68=0 are x=2(2sqrt(2)-5),x=-2(5+2sqrt(2))

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x^{2} + 20 x + 68 = 0

x = 2 (22 - 5), x = - 2 (5 + 22)

Decimal Notation

x = - 4.34314 \dots, x = - 15.65685 \dots

Solution steps

x^{2} + 20 x + 68 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 1 \cdot 68}{2 \cdot 1}

2 0^{2} - 4 \cdot 1 \cdot 68 = 82

x_{1, 2} = \frac{- 20 \pm 8 2}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 20 + 8 2}{2 \cdot 1}, x_{2} = \frac{- 20 - 8 2}{2 \cdot 1}

x = \frac{- 20 + 8 2}{2 \cdot 1} : 2 (22 - 5)

x = \frac{- 20 - 8 2}{2 \cdot 1} : - 2 (5 + 22)

The solutions to the quadratic equation are:

x = 2 (22 - 5), x = - 2 (5 + 22)

f = 0^{2} - 1

x^{2} - x y - 42 y^{2}

4 x - 5 < - 25

\frac{- 5 + 4 i ^{6}}{3 + 3 i ^{33}} + \frac{i ^{387}}{i ^{386}}

- x + 8 = 2 + 3 x + 22

What are the solutions to the equation x^2+20x+68=0 ?
The solutions to the equation x^2+20x+68=0 are x=2(2sqrt(2)-5),x=-2(5+2sqrt(2))
Find the zeros of x^2+20x+68=0
The zeros of x^2+20x+68=0 are x=2(2sqrt(2)-5),x=-2(5+2sqrt(2))